REAL NUMBERS

Exercise: 1.1

1). Express each number as a product of its prime factors:

i). 140                                     iii). 156

iii). 3825                                  iv). 5005

v). 7429

solution:

i). 140

Ans: 140 = 2 X 2 X 5 X 7                                   

ii). 156

Ans: 156 = 2 X 2 X 2 X 2 X 13        

iii). 3825

Ans: 3825 = 3 X 3 X 5 X 3 X 17                                             

iv). 5005

Ans: 5005 = 5 X 7 X 11 X 13          

v). 7429

Ans: 7429 = 17 X 19 X 23   

2). Find the LCM and HCF of the following pairs of integers and verify that LCM X HCF = Product of the two numbers.

i). 26 and 91                                  ii). 510 and 92

iii). 336 and 54          

solution:

i). 26 and 91

26 = 2 X 13

91 = 7 X 13

HCF of 26 and 91 = 13

LCM of 26 and 91 = 2 X 7 X 13

                           =     182

LCM X HCF = 13 X 182 = 2366 . . . . . . . . . . . (I)

Product of the two numbers = 26 X 91 = 2366 . . . . . . . . . (II)

From (I) and (II), we get

LCM X HCF = Product of the two numbers.

ii). 510 and 92

By Prime Factorisation

510 = 2 X 3 X 5 X 17

92 = 2 X 2 X 23

HCF of 510 and 92 = 2

LCM of 510 and 92 = 2² X 3 X 5 X 17 X 23

                           = 23460

LCM X HCF = 2 X 23460 = 46920 . . . . . . . . . . . (I)

Product of the two numbers = 510 X 92 = 46920 . . . . . . . . . (II)

From (I) and (II), we get

LCM X HCF = Product of the two numbers.                          

iii). 336 and 54

336 = 2 X 2 X 2 X 2 X 3 X 7

54 = 2 X 3 X 3 X 3

HCF of 336 and 54 = 2 X 3 = 6

LCM of 336 and 54 = 2⁴ X 3⁴ X 7

                           = 3024

LCM X HCF = 6 X 3024 = 18144 . . . . . . . . . . . (I)

Product of the two numbers = 336 X 54 = 18144 . . . . . . . . (II)

From (I) and (II), we get

LCM X HCF = Product of the two numbers.                          

3). Find the LCM and HCF of the following integers by applying prime factorization method.

i). 12, 15 and 21                             ii). 17, 23 and 29

iii). 8, 9 and 25.

Solution:

i). 12, 15 and 21

12 = 2 X 2 X 3

15 = 3 X 5

21 = 3 X 7

HCF of 12, 15 and 21=  3

LCM of 12, 15 and 21= 2² X 3 X 5 X 7

                                = 420                         

ii). 17, 23 and 29

17 = 1 X 17

23 = 1 X 23

29 = 1 X 29

HCF of 17, 23 and 29=  1

LCM of 17, 23 and 29= 17 X 23 X 29

                                = 11339

iii). 8, 9 and 25.

8 = 2 X 2 X 2 X 1

9 = 3 X 3 X 1

25 = 5 X 5 X 1

HCF of 8, 9 and 25= 1

LCM of 8, 9 and 25 = 2³ X 3² X 5²

                                = 8 X 9 X 25 = 1800

4). Given that HCF (306, 657) = 9, find LCM (306, 657).

LCM X HCF = Product of the two numbers

LCM X 9 = 306 X 657

LCM  = 306 X 65 

             9

LCM = 34 X 65

       = 2210            

5). Check whether 6ⁿ can end with the digit 0 for any natural number n.

Solution: 6ⁿ will end with digit 0 if 5 is one of the primes of 6

Prime factors of 6 = 2 x 3

Since 5 is not a prime factor of 6

Therefore, 6ⁿ cannot end with digit 0.

6). Explain why 7 × 11 × 13 + 13 and 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 are composite numbers.

Let x = 7 X 11 X 13 + 13

       = 13(7 X 11 + 1)

       = 13 (78)

       = 13 X 2 X 3 X 13

       = 2 X 3X 132

Let y = 7 X 6 X 5 X 4 X 3 X 2 X 1 + 5

       =  5 X (7 X 6 X 4 X 3 X 2 X 1 + 1)

       = 5 X (1008 + 1)

       = 5 X 1009

x and y both are expressed as product of primes. Therefore, both are composite numbers.

7). There is a circular path around a sports field. Sonia takes 18 minutes to drive one round of the field, while Ravi takes 12 minutes for the same. Suppose they both start at the same point and at the same time, and go in the same direction. After how many minutes will they meet again at the starting point?

The minimum time when both Sonia and Ravi meet again is the LCM of 18 min and 12 min

18 = 2 X 3 X 3

12 = 2 X 2 X 3

LCM = 2 X 3 X 2 X 3

       = 36

Both Sonia and Ravi meet again after 36 minutes.